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From: slawek on 25 Apr 2010 05:43
I am looking for solutions of this equation:

y'[t] = y[t] - a y[t] y[t-1] y[t-2]

It would be assumed that y[t] =0 for t < 0 .

TIA
slawek

From: A N Niel on 25 Apr 2010 05:51
In article <4bd40edd$0$19181$65785112(a)news.neostrada.pl>, slawek
<slawek(a)host.pl> wrote:

> I am looking for solutions of this equation:
>
> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
>
> It would be assumed that y[t] =0 for t < 0 .
>
> TIA
> slawek
>

then y'[t]=y[t] for 0 <= t < 2 and since y[0]=0 we get y[t]=0
for 0 <= t <= 2 . Continue in this way to get y[t]=0 for all t.
From: William Elliot on 25 Apr 2010 07:00
On Sun, 25 Apr 2010, A N Niel wrote:
> <slawek(a)host.pl> wrote:
>
>> I am looking for solutions of this equation:
>>
>> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
>>
>> It would be assumed that y[t] =0 for t < 0 .
>
> then y'[t]=y[t] for 0 <= t < 2 and since y[0]=0 we get y[t]=0
> for 0 <= t <= 2 . Continue in this way to get y[t]=0 for all t.
>
How do you get that y(0) = 0?
From: William Elliot on 25 Apr 2010 07:50
On Sun, 25 Apr 2010, slawek wrote:

> I am looking for solutions of this equation:
>
> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
>
> It would be assumed that y[t] =0 for t < 0 .
>
There's the trivial solution y(t) = 0 for all t.
For other solutions, it's necessary that 0 isn't in the domain of y
because y'(0) wouldn't exist.

If a /= 0, there's the constant solution
y(t) = k/sqr a, if 0 < t
= 0, if t < 0.

If a = 0, then y(t) = 0, if t < 0
= ke^t if 0 < t.

-- in general
For all t < 2, y'(t) = y(t).
y(t) = 0, if t < 0
= ke^t, if 0 < t < 2

y'(2) = y(2) - a.y(0).y(1).y(2)
shows that y(2) is undefined as y(0) is undefined.

y'(3) = y(3) - a.y(1).y(2).y(3)
shows that y(3) is undefined as y(2) is undefined.

By induction y'(n) for all integers n > 1 is undefined.


For 0 < t < 1,
y'(2 + t) = y(2 + t) - a.y(t).y(1 + t).y(2 + t)
= y(2 + t) - aek^2.e^2t.y(2 + t)
= y(2 + t) - aek^2.e^2t.y(2 + t)
= y(2 + t)(1 - aek^2.e^2t)
= y(2 + t)(1 - ae^-3.k^2.e^2(t + 2))
is separable equation.

y'(2 + t)/y(2 + t) = 1 - ae^-3.k^2.e^2(t + 2)
log y(2 + t) = t + c - ak^2 / 2e^3 * e^2(t + 2)

Using that solution one can now consider

y'(3 + t) = y(3 + t) - a.y(1 + t).y(2 + t).y(3 + t)

for 0 < t < 1, and by induction, if you can yuck through the muck,
find for each n > 1, peicewise solutions for y(n + t), 0 < t < n.
From: A N Niel on 25 Apr 2010 08:01
In article <20100425035622.G21838(a)agora.rdrop.com>, William Elliot
<marsh(a)rdrop.remove.com> wrote:

> On Sun, 25 Apr 2010, A N Niel wrote:
> > <slawek(a)host.pl> wrote:
> >
> >> I am looking for solutions of this equation:
> >>
> >> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
> >>
> >> It would be assumed that y[t] =0 for t < 0 .
> >
> > then y'[t]=y[t] for 0 <= t < 2 and since y[0]=0 we get y[t]=0
> > for 0 <= t <= 2 . Continue in this way to get y[t]=0 for all t.
> >
> How do you get that y(0) = 0?

y differentiable => y continuous
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